UPSC Maths 2025 Paper 2 Q5b — Solution

10 marks · Section B

Question

Solve the following system of linear equations by Gauss-Seidel method: $10x + 2y + z = 9$ $2x + 20y - 2z = -44$ $-2x + 3y + 10z = 22$

Technique

Apply the Gauss-Seidel iterative method. The coefficient matrix is diagonally dominant, so the iteration converges; rearrange each equation for its diagonal variable and use the most recent values within each sweep.

Solution

Check diagonal dominance.

Row 1: $|10| \ge |2| + |1| = 3.$ ✓
Row 2: $|20| \ge |2| + |-2| = 4.$ ✓
Row 3: $|10| \ge |-2| + |3| = 5.$ ✓

The system is strictly diagonally dominant, so Gauss-Seidel converges for any starting guess.

Iteration formulas. $x^{(k+1)} = \frac{1}{10}\left(9 - 2y^{(k)} - z^{(k)}\right),$ $y^{(k+1)} = \frac{1}{20}\left(-44 - 2x^{(k+1)} + 2z^{(k)}\right),$ $z^{(k+1)} = \frac{1}{10}\left(22 + 2x^{(k+1)} - 3y^{(k+1)}\right).$

Start with $x^{(0)} = y^{(0)} = z^{(0)} = 0$ .

iter $k$	$x$	$y$	$z$
1	$0.900000$	$-2.290000$	$3.067000$
2	$1.051300$	$-1.998430$	$3.009789$
3	$0.998707$	$-1.998892$	$2.999409$
4	$0.999837$	$-2.000043$	$2.999980$
5	$1.000011$	$-2.000003$	$3.000003$
6	$1.000000$	$-2.000000$	$3.000000$

Sample first sweep: $x^{(1)} = \tfrac{1}{10}(9 - 0 - 0) = 0.9,$ $y^{(1)} = \tfrac{1}{20}(-44 - 2(0.9) + 0) = \tfrac{1}{20}(-45.8) = -2.29,$ $z^{(1)} = \tfrac{1}{10}(22 + 2(0.9) - 3(-2.29)) = \tfrac{1}{10}(30.67) = 3.067.$

The iterates stabilise to $(1, -2, 3)$ by the 6th sweep (4-decimal accuracy already by iteration 4–5).

Answer

$\boxed{\,x = 1,\quad y = -2,\quad z = 3\,}$